Suppose I have two charged capacitor plates that both are isolated and carry a charge density $D = \frac QA$. According to textbook physics the electric field between them is given by $E=\frac D {\epsilon\epsilon_0}$ and the voltage by $U = Ed = \frac {Dd}{\epsilon\epsilon_0}$ with $d$ the distance between the plates. According to the formula for the voltage from above I could set any voltage between the plates if I just separate them far enough from each other and also the electric field would be constant no matter how far the plates are apart which is also quite counter-intuitive. As far as I remember this is true as long as $d$ is small compared to the size of the charged plates.

But what if this condition no longer holds? What is happening then? Is there another formula for this case that is comparably simple? I would suppose that for very large $d$ the whole thing can be seen as two point charges which would give a $\frac1r$ dependency of the voltage. But what is happening in between?

2 Answers
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Here is a simplified approach to this question-I hope it is not too simplistic. Sorry I could not upload the mathematics and the illustrating diagram from my computer file. I need to learn how to do this, or I would appreciate if someone could leave some ideas.

Basically the approach by "MyUserIsThis" is intuitively sound. The analysis is not detailed enough to show how V depends on d (distance between the plates) at small and large d.

Imagine the two parallel plates $P_1$ and $P_2$ with finite Area $A_1$ and $A_2$, carrying eletric charges with uniform densities $D_1$ (charge $+Q_1$) and $D_2$ (charge $-Q_2$) respectively. The plates are placed on top of each other ($P_2$ above $P_1$). We assume uniform densities for simplicity. Now, choose two differential elements: $dA_1 = dx_1dy_1$ on $P_1$ at point ($x_1, y_1, 0$) and $dA_2=dx_2dy_2$ on $P_2$ at point ($x_2, y_2, d$). The potential difference between the plates is given from standard electrostatic theory, leading to the following general but ‘complicated’ double integral on the surfaces $A_1$ and $A_2$

Therefore, the potential difference drops as $1/d$, which is equivalent to saying that for large $d$, the two plates see each other as point particles of charge $+Q_1$ and $-Q_2$, as mentioned in the previous answer. I hope this adds some clarity to the answer.

Thanks Benedikt. I appreciate your editing the mathematics of my answer. I hope soon I shall be able to do this myself.
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JKLFeb 2 '13 at 0:08

You're very welcome. You will see, writing formula isn't that complicated as in most cases you will need only a few basic commands and the syntax is nearly completely derived from LaTeX syntax. Therefore also a quick look in to some short introduction on maths in LaTeX could be helpful. Also thanks for the 'complicated' double integral formula, maybe I can reduce it a it in complexity for my special use case. Oh, nearly forgot: if you write an @name into your comment, user "name" will be notified that there is a new comment
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Benedikt BauerFeb 2 '13 at 18:24

Comparably simple, no. That condition is there so you can avoid boundary effects in the limits of the sheets. If they're close enough you can approximate their size/distance ratio to infinity and this allows you to suppose the field /by Gauss law) has tu be uniform inside. This is what actually happens:

As you can see the field stops being uniform an perpendicular to the sheet in the borders.

That condition may not be true always, by the way. For example, if you have a couple of plates and you connect them to a battery, then the battery will maintain the voltage difference between the plates, even if you change the distande $d$, by moving charges in and out, until the potential bewtween the plates is the same as in the battery and you get to a stationary point.
If instead of connecting a battery you charge them with an amount of charge $\pm q$ on each plate, then what happens is what you have said. By using Gauss law, you can create gaussian surfaces in which to integrate the field and deduce that if the charge is constant, then the field $E$ will be constant, and the potential is what's going to change. This are the gaussian sufaces you could pick (the rectangular pillbox):

If the field is constant and uniform, the potential will be, by definition:
$$V=\int_0^d Edx=E\int_0^ddx=Ed$$
And that's the way you get your formula.
All of the above can be applied as long as we keep the approximation of very large and close plates, so we can ignore boundary effects.

Thank you for the elaborate explanation. Unfortunately this doesn't really answer my question. I already had a clue how to come to the simplified formula but I don't know how to calculate this if the simplified formula is not applicable.
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Benedikt BauerFeb 1 '13 at 11:32

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@BenediktBauer You can't get a general formula in the case in which you can't ignore boundary effects. When you need values of field or voltage in complex cases like that one, you generally use numerical methods (computers) to get numerical values, but you can't get formulas. That's how pictures like the first one in my answer are done, by numerical approximations, that with computers can have huge preccisions.
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MyUserIsThisFeb 1 '13 at 12:06